Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-2x+y &= -4 \\ -5x+y &= -8\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-5x = -y-8$ Divide both sides by $-5$ to isolate $x$ $x = {\dfrac{1}{5}y + \dfrac{8}{5}}$ Substitute this expression for $x$ in the first equation. $-2({\dfrac{1}{5}y + \dfrac{8}{5}}) + y = -4$ $-\dfrac{2}{5}y - \dfrac{16}{5} + y = -4$ Simplify by combining terms, then solve for $y$ $\dfrac{3}{5}y - \dfrac{16}{5} = -4$ $\dfrac{3}{5}y = -\dfrac{4}{5}$ $y = -\dfrac{4}{3}$ Substitute $-\dfrac{4}{3}$ for $y$ in the top equation. $-2x- \dfrac{4}{3} = -4$ $-2x-\dfrac{4}{3} = -4$ $-2x = -\dfrac{8}{3}$ $x = \dfrac{4}{3}$ The solution is $\enspace x = \dfrac{4}{3}, \enspace y = -\dfrac{4}{3}$.